$g(t) = 4t^{3}+6t^{2}+5t+2-3(f(t))$ $f(n) = -6n^{2}$ $ g(f(0)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = -6(0^{2})$ $f(0) = 0$ Now we know that $f(0) = 0$ . Let's solve for $g(f(0))$ , which is $g(0)$ $g(0) = 4(0^{3})+6(0^{2})+(5)(0)+2-3(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = -6(0^{2})$ $f(0) = 0$ That means $g(0) = 4(0^{3})+6(0^{2})+(5)(0)+2+(-3)(0)$ $g(0) = 2$